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2 years ago

A 2.5 L container is filled with propane. The ambient temperature is 25°C and the

Chemistry

2 answers:

vlabodo [156]2 years ago

6 0

Answer:

3.3atm is the new pressure of the container

Explanation:

In the car, the temperature of the container increases from 25°C =298K to 55°C = 328K. To solve this question we must use the Boyle's law that state that the pressure of a gas is directly proportional to its absolute temperature under constant volume. The equation is:

P1T2 = P2T1

Where P is pressure and T absolute temperature of 1, initial state of the gas and 2, final state.

Replacing:

P1 = 3atm

T2 = 328K

P2 = ?

T1 = 298K

3atm*328K = P2*298K

P2 = 3.3atm is the new pressure of the container

egoroff_w [7]2 years ago

4 0

Answer:

3.3 atm

Explanation:

First we convert 25 °C and 55 °C to K:

25 °C + 273.16 = 298.16 K

55 °C + 273.16 = 328.16 K

We can solve this problem by using Gay Lussac's law, which states that at constant volume:

T₁P₂=T₂P₁

Where in this case:

T₁ = 298.16 K

P₂ = ?

T₂ = 328.16 K

P₁ = 3 atm

We input the data:

298.16 K * P₂ = 328.16 K * 3 atm

And solve for P₂:

P₂ = 3.3 atm

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An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

4 0

3 years ago

Which of the following are acids by the Bronsted-Lowry Definition:

olga55 [171]

ANSWER IS (A)

EXPLANATION:

Bronsted-Lowry concept states that a substance is an acid if it can act as a H+ donor.

HCl in aqueous solution means that HCl is present in water, HCl + H2O --> H3O+ + Cl-. This reaction will take place, the H+ from HCl will be donated to H2O. So, HCl is a bronsted-lowry acid by definition.

However, Methanol (CH3OH) its written that it is liquid, i.e. pure methanol, CH3OH(l). It is both acidic as well as basic. when it is mixed with water then it behaves as an acid.

The last one ammonia in gas phase is also neutral because its not in water. if mixed in water it behaves as a base.

6 0

3 years ago

Convert 25.4 grams of barium phosphate, Ba3(PO4)2 to formula units.

Salsk061 [2.6K]

To solve this questions you first need to find the number of moles of barium phosphate you have. The molar mass of barium phosphate is 601.93g/mol.
24.4/601.83 = 0.0402 moles barium phosphate
Then you need to use avagadro’s number, 6.022 x 10^23, which is the number of molecules or formula units in a mole.

6.022 x 10^23 * 0.0402 = 2.42 x 10^22 formula units

7 0

2 years ago

2. What types of organisms have cells with very large vacuoles?

Sedaia [141]

The answer would be plant cells

5 0

3 years ago

A bowling ball rests on the floor. The bowling ball is given a push.

lyudmila [28]

Answer:

A. It will move.

8 0

2 years ago